Problem: What is the value of $\dfrac{d}{dx}(2x^2-6x+5)$ at $x=-4$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-10$ (Choice B) B $-14$ (Choice C) C $61$ (Choice D) D $-22$
Let's first find the expression for $\dfrac{d}{dx}(2x^2-6x+5)$ and then evaluate it at $x=-4$. According to the sum rule, the derivative of $2x^2-6x+5$ is the sum of the derivatives of $2x^2$, $-6x$, and $5$. The derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ For example, this is the derivative of the first term: $\begin{aligned}\dfrac{d}{dx}(2x^2)&=2\dfrac{d}{dx}(x^2)&&\gray{\text{Constant multiple rule}}\\\\ &=2\cdot (2x^1)&&\gray{\text{Power rule}}\\ \\ &=4x\end{aligned}$ Here is the complete differentiation process: $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(2x^2-6x+5) \\\\ &=2\dfrac{d}{dx}(x^2)-6\dfrac{d}{dx}(x)+\dfrac{d}{dx}(5)&&\gray{\text{Basic differentiation rules}} \\\\ &=2\dfrac{d}{dx}(x^2)-6\dfrac{d}{dx}(x)+0&&\gray{\text{Constant rule}} \\\\ &=2\cdot 2x-6\cdot1x^0&&\gray{\text{The power rule}} \\\\ &=4x-6 \end{aligned}$ So we found that $\dfrac{d}{dx}(2x^2-6x+5)=4x-6$. Plugging in $x=-4$ and evaluating using the calculator, we find that the value is $-22$. In conclusion, the value of $\dfrac{d}{dx}(2x^2-6x+5)$ at $x=-4$ is $-22$.